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How do wing ribs withstand lateral lift force?
During roll,there is a vertical component of lift that keeps the plane in the air and an horizontal component that actually rolls the plane,how are wing ribs designed to help withstand this vertical(now sideways to ribs) lift force since wing ribs are inherently weak laterally?

TL;DR: from the ribs perspective nothing changes.
Look at the three figures you posted: the lift is always straight in the middle of the aircraft. Tilt the aircraft, the lift tilts with the aircraft.
From the perspective of the ribs there is no side force, the lift is still perpendicular to the wings (actually, to the incoming airflow, but is a good approximation).
The only case where the wings will see some side force is when the aircraft is in a sideslip, i.e. $\beta \neq 0$.
In such a situation is not the ribs that carry the side load, but the wing beams, that run along the whole wing.


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Can I use PVC pipes for the rib on the wings of my homemade airplane?

Can I use a PVC pipe as the rib of the wings of my homemade airplane?A rib is already load tested, and it failed at 40kg of weight and it is supposed to hold just 10kg in flight.

I don't have an answer.  Is the plane your own design? Are you following someone else's plans and making mods to it?

The fact that you do not specify what kind of load it was and give it in kg makes me uneasy about this design and its test pilot.

Have you considered joining EAA? They're got all kinds of resources for homebuilt projects. Not sure this can really be answered since we won't know all the specifics of your design. We wouldn't want to say, "sure, go ahead" and have you get injured or killed if it fails.

@Federico.......The plane has 14 ribs and each  rib should hold 10kg (100N of force) each...plus the plane should weigh about 60kg and the Test pilot 55kg+

@crossroad......its my own design

@DavidTeahay without knowing a lot more than that, it's really hard to tell. You'd need to give a wing loading and expected total weight to even start sizing the torque at the wing root, which is what will decide if your tubes are enough, not the force they resist on their own in some unspecified direction and under unknown boundary conditions.

Total weight is 60 + 55 == 115kg.  Total strength is 14 x 10 = 140kg.
That's about 121%.    The absolute bare minimum should be around 225% to account for load-factor and safety margin.

https://aviation.stackexchange.com/questions/46965/what-physical-tests-are-involved-with-verifying-a-wing-spar-is-suitably-construc Look at Peter Kämpf's answer to see how a wing is load-tested. They are load-tested that way to mimic the forces and moments experienced during flight. From the picture you've posted, it seems you haven't considered those forces/moments.

I would agree with @TomMcW that joining EAA would probably be a good idea. You could also join the Homebuilt Airplanes Forum.
I assume you already know the basics of choosing materials. They are a balance of strength, weight, and cost. I assume your choice of PVC is low cost. So, the big question would be whether it has the rest of the required properties.
I assume that you figured wing loading like, say, 5 lbs per square ft or 0.2 kg per square meter. Each rib has to handle the load up to halfway between the other ribs. An Aeronca C-2 has ribs about 12 inches (30.5 centimeters) apart which is closely spaced and it is a fairly light airplane at 700 lbs (318 kg) gross weight. So, we could roughly divide the gross weight by wingspan. However, the actual load on the wing isn't even. It is greater inboard near the fuselage and then falls as we go outward to the wingtip. So, multiply this by 1.4 for the greatest load at the root. Then we multiply this by the design gee rating. Let's use the Federal Aviation Administration (FAA) utility rating of 4.4 gees. Then we have to multiply by the safety factor which is normally 1.5. So, let's do the math for the C-2. We'll load it to less than gross with an average 160 lb (72.5 kg) pilot plus 9 gallons (34 liters) of fuel.
620 lbs (282 kg) gross weight
36 foot (11 meter) wingspan
12 inch (30.5 centimeter) rib spacing
620 lbs (282 kg) / 36 = 17.2 lbs (7.83 kg) per rib
17.2 x 1.4 = 24.1 lbs (11 kg) on the inboard ribs
24.1 x 4.4 gees = 106.1 lbs (48.3 kg) under highest maneuvering load
106.1 x 1.5 safety factor = 159 lbs (72 kg) per rib breaking strength
At this point I'm wondering how you got a maximum load as low as 10 kilograms unless you are talking about a large, remote control airplane.
In case you are talking about something that is intended to carry a human, I'll continue. It might be tempting to use the lower, 3.8 gee normal rating instead of the 4.4 utility rating. However, aircraft with lower wing loading are more susceptible to wind gusts so I don't think this would be a good idea.
To test the rib lift load, you space the weight with the center at the quarter chord position rather than evenly spaced across the chord. We also need to account for the pitching moment, but this is a little trickier. We know that:
dynamic pressure x wing area x coefficient of lift = lift
and
dynamic pressure x wing area x coefficient of pitch x chord = pitching moment
therefore
lift / lift_coef x pitch_coef x chord = pitching moment
lift = 159 lbs (72 kg)
For a Clark Y airfoil at 5 degrees:
lift_coef = 1
pitch_coef = 0.08
let's say the chord is 4.3 ft (1.3 meters)
The conversion is simple in pounds but metric requires multiplying by the acceleration of gravity, 9.834 meters /sec^2 since Newtons are not Earth relative whereas pounds are.
159 lbs (72.12 kg x 9.834) / 1 x 0.08 x 4.3 ft (1.31 meters) = 55 lb ft (74.57 newton meters)
55 lb ft / 159 lbs = 0.35 ft
(74.57 newton meters / 9.834 meters / sec^2) / 72.12 kg = 0.105 meters
So, we see that to match the torque of the pitching moment, we'll have to move the center of the test weight rearward 0.35 ft (4.2 inches) or 0.105 meters (10.5 cm).
If your rib is strong enough at this point then you still need to ask if sunlight would breakdown the plastic and make it brittle, if it might be too brittle in cold weather, if you would be able to inspect it.
I have seen PVC used in place of bamboo to make a cheap hang-glider. However, these were not expected to fly very high and since they used removable sails, inspection was not an issue.

Yes, of course you can.
However, it's probably not a very good idea.
PVC pipes are pretty strong and light; circular-section PVC pipes have an excellent strength-to-weight ratio.
Being strong and light isn't enough though. Construction materials need to have the appropriate properties (like being strong and light) in the right conditions.
In the case of an aeroplane wing, that includes the conditions that aeroplane wings are subjected to: rapidly-changing stresses, vibration, sustained load, torsion and so on. You need to know: will PVC remain strong under such conditions?
Then presumably, you will need to find various ways of joining your PVC sections to each other and to other materials. How will these joins be made? How will they behave under the conditions an aeroplane wing is subjected to? How will the joining methods affect the properties of the PVC pipes?
Aircraft designers tend towards the conservative; even the most radical of aircraft designers would not be considered daring compared to the radical leaps taken by designers in other fields. That's because the chief reward for brave leaps in aviation is death.
All the same, over its history, aviation has tried all kinds of things (usually, quite carefully). If aviation has successfully used PVC pipes in aircraft construction, there'll be a record of it and a tradition of it. The successful pioneers of PVC-pipe-based aviation will not, however, have embarked on their adventures on the basis of an encouraging Stack Exchange reply.
If you don't already know the answers to the questions I mentioned above, and are not already in possession of reliable information about techniques, tools, practices and theory for working safely with PVC load-bearing structures, I'd prefer to use PVC pipes for plumbing only.

Analyze this problem in terms of structure dynamics.  A structural member of circular cross section is isotropic with respect to stiffness; that is to say it resists bending equally from all directions.  This is not the best choice for an airframe.  Because the lifting members of an airframe need to resist force primarily along the vertical axis, the underlying structure of those members should be anisoptropic.  An example of an anisotropic cross section is an 'I-beam'.  That kind of section yields most bending resistance per pound of material.

Theoretically, you are correct. Practically, isotropic design is helpful is the load cases cover all kinds of loads. A tube is much better protected against buckling under compression than the I-beam, for example.

@DavidTeahay We're referring to EAA (Experimental Aircraft Association), not FAA. They don't have a chapter in Nigeria, the closest is South Africa. But I'm sure they have resources you can access through the internet or mail. The biggest part of their existence is to help people who are building their own aircraft, so I'm sure they could hook you up with advice.

Oh EAA....Okay I'll reach out to them for their resources....thanks a lot @TomMcW

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Would the Wright Flyer have flown better with thicker airfoils?
The Wright Flyer(s) flew with very little wing thickness,would they have flown higher, faster and longer if the wings had greater thickness?
Secondly, how do thickness and camber affect a  wing's performance?

Would the Wright Flyer have flown better with thicker airfoils?

The Wright Flyer: Probably no. Most other aircraft: Definitely yes.
Thin airfoils are best at one specific angle of attack and their drag increases rapidly if you move away from this angle in both directions. The small leading edge radius will only let the air negotiate the leading edge properly when the stagnation point is right in the center of the leading edge, which happens at the design angle of attack. Then a thin airfoil works really well - that is why slotted flaps should be thin: Their flow is directed by the wing ahead of them.
The Wright Flyer I had just enough power to stay airborne at its lowest drag speed, so it flew at just one angle of attack. Of course, the Wright brothers had too little knowledge of aerodynamics to select just the right camber to place the design point of the airfoil into its operating point. The camber on the Flyer I was probably too high, because in later airplanes the Wrights reduced airfoil camber. Not a thicker, but a different airfoil could most likely have shown better results.
Why did the Wrights use a thin airfoil, anyway? They followed the advice of Otto Lilienthal who designed his airfoils from cross sections of bird wings. Birds can vary their wing camber in flight, so they adjust the wing such that the stagnation point always sits in its optimum position. Lilienthal even discovered that thick airfoils performed better but did not understand why and preferred to follow Mother Nature in his airfoils.
From Fred E. C. Culick's AlAA Wright Flyer Project - Report WF 84/09-1:

So the Wrights followed Lilienthal and used thin, highly cambered
  airfoils resembling the cross-sections of birds' wings. They were
  misled to believe that airfoils of that sort produced the highest
  ratio of lift/drag. There is in fact much truth in this conclusion if
  data are taken for small wings at the low speeds the Wrights used in
  their wind tunnel tests.

The two main advantages of a thick airfoil are

the higher flexibility it gives in various flight conditions.
the structural advantage of using an internal spar and doing away with the wire bracing, even in bi- and triplanes.

Airfoil thickness comparison by D.R. Kirk, Florida Institute of Technology (picture source)
Just to disprove the misconception that thin airfoils have a poor L/D: They haven't, at least in a small angle of attack region. Below are the L/D and lift curve slope plots of the Gö 417 airfoil, something like the lab mouse of thin airfoils. To achieve a lift coefficient of nearly 1.6 at a Reynolds number of 1 million is quite an achievement, and doing so at that low drag doubly so.

Source: Airfoiltools.com

There were several other limitations on the first Flyer besides the airfoil -- the engine weighed four times what the brothers had been promised it would (because the block was cast, rather than fabricated from sheet), and would have been underpowered by today's standards even if it had delivered to spec.  The whole machine came in just under 300 kg; a modern ultralight that size would weigh less than half as much and have nearly twice the power.
Given those limitations, the airfoil is a relatively minor consideration.  Yes, there are modern airfoils that would perform better in terms of coefficient and lift-to-drag -- but the thing that would have helped the most would have been saving 40 or so kg. on the engine weight.
It's worth noting that through the First World War, with aircraft flying at four or five times the speed of the first Flyer, at altitudes up to 4000+ m, inverted and other aerobatic maneuvers, what we'd call relative thin, cambered airfoils persisted.  Look at the wing section of a Sopwith Camel -- though slightly thicker and more deeply cambered than the original Flyer's airfoil, it was still thinner than anything you'll find today except on single-surface hang gliders and ultralights.
How thickness and camber affect flight characteristics is subject matter for a full length college level textbook, but short version: more camber tends to increase lift coefficient over some range of angle of attack, but increases drag in proportion; adding thickness increased coefficient over a broader AoA range, with less drag increase, but has less effect at low AoA than increasing camber.  There's a compromise, with both increased by similar amounts, that includes the well-known Clark Y, NACA 6409 and 6412, and many other common airfoils used in general aviation, especially when wood and fabric were the building materials.

I recall being told that the persistence of thin airfoils during WW1 was due to an incomplete understanding of the effects of the Reynolds number, which led to skewed wind tunnel results. I don't see any questions about it on the site, so it may be worth asking.

My point was that, with the power and weight of the original Flyer, it would have been a marginal performer even with a modern airfoil.  Like trying to fly a Cub with a garden tiller engine.

I see, I just interpreted the passage about the Great War as proof that the airfoils were just fine, which to me seems the core of OP's question. That the powerplant was a bigger issue is relevant, but not directly asked about.

Bad airfoils + much better power + more pilot knowledge = far higher performance.

The Dr.I airfoil was rather thick and allowed to do away with wire bracing. This gave the Dr.I a decisive advantage, given its engine's low power.

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Are sudden turns and descents a standard emergency procedure?
First, I'm not looking for speculation on an accident, I just want to know if pilots are trained to make certain maneuvers in response to an emergency.
In reading about the jet that just went down in Egypt, the story stated

"It turned 90 degrees left and then a 360 degree turn toward the right, dropping from 38,000 to 15,000 feet and then it was lost at about 10,000 feet," Kammenos added.

Assuming the report is accurate, are those maneuvers a standard procedure of some kind? Are pilots trained to maneuver like that in an emergency? If so, in what type of emergency would those maneuvers be useful?

And to add on to @Federico we won't really know the exact details until they have read the data from the "Black Boxes"

I agree with @federico, although I'd add that I know of no "normal" reason for these manoeuvres. But we already know this is an emergency situation of some kind, so "normal" doesn't apply

At the bottom of the article there's a quote CNN aviation analyst Les Abend: The 360-degree turn, that seems very abrupt. It's not something I would do in any major emergency unless I was losing control of the aircraft.

Just wanted to point out, excellent job keeping this question generally applicable and far away from speculation.  Well asked.

@JayCarr I would say well edited

@Federico And probably well commented to get to those edits.  So, good job on everyone :D.

What makes you think that those were, in fact, pilot maneuvers? There is no indication of that in the article you linked. All that the article says is that the plane moved like that. It could just as well be the exploded, destroyed airframe merely plunging through the air. Or the first turn is the pilots trying to avoid a plane on a collision course, and the rest is what happened to the airplane after the collision ripped one wing off? Or … (insert about a 1000 other possibilities here that don't involve the pilots maneuvering).

@Jörg W Mittag I didn't know if they were pilot maneuvers. That is exactly what prompted the question "Were they pilot maneuvers?" and the continuation "Do emergency pilot maneuvers even exist?". The update later on in the article from the CNN Aviation Analyst hadn't been published when I first posted the question.
Thanks for all the help editing the question, all!

@TomMcW Personally, I wouldn't normally encourage anyone to take the word of a CNN "aviation analyst/expert," as most of the ones I've heard on there have had only little clue about what they were talking about (and not just for aviation, but any technical subject.) However, the wording in this particular case suggests he's actually a pilot, though, so that might be a bit more reliable than CNN's normal "aviation experts."

@reirab Yeah. I could probably qualify as an "aviation analyst" for CNN! The guy they quoted has good creds, though. He's a 777 pilot and has a column in Flying magazine.

There are no emergency procedures I'm aware of that requires a pilot to turn left 90 degrees and then turn right 360 degrees.
There is a procedure for situations requiring rapid descent for things like medical emergencies and fires.  They consist of a 45 degree banked, full-flap, power-idle descent flown at maximum flap extension speed.  So it is a spiraling-turn descent.
Also, the Cessna 172N checklist instructs the pilot to slip in the opposite direction from a wing fire to keep the flames blown in an outboard direction.

Thank you! Just out of curiosity -- what is the approximate rate of descent with this procedure on an 'average' commercial airliner? This question says that normal rates are probably under 4,000 fpm -- how much faster is this procedure?

@jEsp I found this generic Boeing checklist. It looks like they don't utilize a bank like smaller aircraft do.  I speculate they can descend so fast as it is, that increasing the descent rate may begin to reduce safety so that it outweighs the benefit.

There is no such procedure for making these turns, however you can make orbits to lose altitude which is basically 360 turns. You may need it when you are very high and need to lose altitude. Let say you lost engines, executing powerless flight and you have very good option for airport just below you. So you can come on top the airport and begin orbiting. Also in military there is pill off manoeuvre making 360 to land.
For this crash, I think there is a total loss of control associated with electrical fire after examining ACARS messages. You can find the link here :
http://avherald.com/h?article=4987fb09&opt=7168
May be detailed but nice to read all.
Bests,

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Is there a site that tracks all flights?
Some of my flights are tracked on sites like Flightradar24 and others are not, regardless of flight plans being opened and closed. It would be nice to go to a website that tracks all flights, so I can review the flight dynamics after I land.
I am interested in a website that provides this service, not an app (I have had limited success with these).

Clarification: You want to view every single flight that occurs or do you want to view the full flight?

Are you talking about IFR only, or both IFR and VFR flight plans?

@SMSvonderTann yes I would like to see all flights

@JonathanWalters both would be great, however VFR would prove to be more useful

My understanding and experience is that FAA VFR flight plans are never tracked, only IFR flight plans. VFR radar flight following radar positions may sometimes show up, but unreliably. This seems to be confirmed by the FlightAware.com FAQs

I'm not totally sure what you're asking for but if you want to do detailed analysis of your own flights, you need more precise position data than you can get from an online tracking site. That means recording GPS flightpath data and then visualizing it, which is probably going to require an app of some sort. FWIW, I've had good results using a Garmin Virb to capture the data and Cloudahoy to analyze it.

No, there will always be flights that are private, or for any other reason not traceable.

Your answer is correct in the implication that there is currently a means of opting out of flight tracking. However, your assertion that there will always be flights a that are not traceable is baseless. Indeed, there have been attempts at eliminating that option in the not so recent past. And, perhaps more importantly, ADS-B may render private, untraceable flights more or less a thing of the past.

@JonathanWalters Tell that to the guy flying an F-22... ADS-B doesn't help much if the transponder is turned off and the aircraft is designed to not show up on radar. :) Maybe "always" is a stretch, but "for the foreseeable future" would be accurate enough.

Tracking sites use FAA data for the most part, augmented by ADS-B data nowadays.
For FAA data, they use what's in the National system. This is usually 99%+ of IFR flight plans, and some VFR flight following flight plans. IFR flights plans that aren't included are those that are local area pop-up requests(staying with the same approach control with a pop-up IFR clearance).
For VFR flights, the only data in the NAS is what's input by controllers for flight following that leaves their airspace(leaving a TRACON headed to a center or another facility, usually; sometimes you can get a NAS code for just a local flight). VFR flight plans sent to and activated by flight service are not sent to ATC, all they are is a tool for rescuers to search for you if you're late and never check in and a search needs to go find you.

Sometimes (generally it seems to be if your trip crosses center boundaries) VFR flight following will get entered into the national system rather than just local codes, and you might show up on services like FlightAware. this is pretty unpredictable though so you shouldn't rely on it.

The short answer to this question is to fly ADS-B. If you use ADS-B, then in general, you will always be tracked unless you employ a specific blocking mechanism.
Other than that your only recourse is to always fly under a IFR flight plan which should result in tracking 98% of the time.
If you are not getting tracked, it may be because you are using a VFR flight plan, or are flying in an area outside Europe or the United States.


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When do you contact ATC when departing from a non-towered airport into class D controlled airspace?
I had this question posed to me the other day and I wanted to make sure I responded correctly.
When do you contact ATC when departing from a non-towered airport into class D controlled airspace?
This is the situation where the uncontrolled airport sits under the Class D shelf.

Class D is usually a cylinder so there's no shelf to be underneath; I'm assuming you're asking about departing from an uncontrolled airport that's actually in class D, like this one (N96):

Because class D goes to the surface I'd try to contact ATC on the ground before departure if possible, either on a regular frequency or via an RCO. And it would be worth checking the Chart Supplement for any local rules.
Otherwise, 14 CFR 91.129 says you should contact ATC as soon as you can after departure:

(2) Departing flight. Each person—
[...]
(ii) From a satellite airport without an operating control tower, must
  establish and maintain two-way radio communications with the ATC
  facility having jurisdiction over the Class D airspace area as soon as
  practicable after departing.

There are some special cases like KDTS where you can't takeoff without contacting ATC first, but it's in a special rules area.


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Is there a relation between material choice and produced lift at rotation?
I'm currently learning about Rotary Winged Aircraft, and I have a question regarding lift and fluid flow.
Given two wings of different material but same shape, will there be a difference in the amount of lift, and the amount of air that is pushed down? Is the lift created related to the material choice in any way?

In theory no effect at all if the blade shape and surface finish are the same. But bear in mind if the material and/or construction affect the blade flex and twist, there may be differences in the lift. (I gather twist of the retreating blade is important in helicopter design too by the way. Maybe someone will address this.)

You may want to use different materials specifically for their surface finish, but unless you have some very special materials, the effect wouldn't be too big.

Also, can you be more specific regarding whether you are talking about an ideal rotor blade or an actual one? In particular, blades of different material will react differently to vibration, "droop" (on startup/shutdown), and inertia (in autorotations), and generate differently amounts of lift.  In an ideal blade, I don't think material is, well, material.

@rbp There would be no difference of lift generated by a fabric covered blade, compared to an aluminum one, further compared to carbon fiber? I'm not sure I agree that material is not relevant.

yes, that's exactly my point: if its an ACTUAL blade or an IDEAL one. The ACTUAL blade cares, the IDEAL one doesn't

Yes.
Some materials are easier to shape (think composites versus metal) or stay in shape better over time or under stress (think metal versus wood). This difference between the intended and the actual shape will have the biggest material-related impact on lift.
If you slightly change the question and ask for useful lift (lift minus weight of the structure), the stiffness and relative strength of the specific material must be added to the comparison. Useful parameters are the ultimate strength relative to density (breaking length) or stiffness relative to density (strain length). A stiff structure deforms less under load, so its shape will change less. Ideally, the structure is shaped such that only under load it assumes the aerodynamically most advantageous shape while its elasticity helps it to avoid being overloaded (aeroelastic tailoring).
The surface finish is also material-related, but is more affected by the manufacturing method. A rough surface is better at very low flow speeds while a smooth surface is better at moderate to high flow speeds.


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What should I do to become a commercial pilot?
I am 25 years old, I have completed my bachelor  degree  from electrical  engineering. So what should I do to become a pilot?

Welcome to the site! You might want to look at our existing career questions and it would be good to know which country you're in. It's often difficult to answer career questions because there are so many personal, financial and legal variables, so the more specific you can be, the better.

I am from Bangladesh,  I am asking here that, do I eligible for that? what are the main factor to become commercial  pilot.

The main factor in becoming a pilot is.... Pilot training! You need to find a good local flying school. I have found 4 flying schools in Bangladesh already by doing a Google search.

If you plan to train and fly in the US...
You should spend that money on an MSE or MBA, get your private/instrument, and then buy your own plane and fly whenever you want.
However, if you feel you must become a commercial pilot you should find a local flight school and lay out a plan with someone who has authourity to negotiate pricing.
-lay out your objective.  You are paying for a private license, instrument, all of the time building for commercial, and then commercial.  That's a minimum of 250 flight hours plus a lot of billable hours for a very lucky CFI.  You can negotiate an awesome rate if you guarantee the school that business in s short amount of time.
-DO NOT pre-pay more than $1000 at a time.  Keeping funds on account is fine, just don't give them more than you're comfortable losing if suddenly they close.  It happens.
-stick to a schedule.  Fly a lot.  It's cheaper to knock it all out at once.  You have to relearn stuff if you only fly once or twice a week.
-don't get caught up comparing part 61 and part 141 schools.  Just go with the most affordable option.
-research airline minimums.  A commercial cert won't necessarily get you a job.  If fact, a commercial cert means very little wothout hours to go with it.  Think about adding a CFI cert to your shopping list.


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When do you contact ATC when departing from a non-towered airport into class C controlled airspace?
When you're departing from an airport within controlled airspace, when should you contact ATC?  This is when the airport you were departing from is non-towered.
Scenario: I am departing a non-towered airport in class Charlie airspace. I need to request VFR flight following to another airport. When do I contact ATC?

If you can't contact them on the ground, as soon as practicable after take-off.

Be sure to contact them though, this is not optional... You are flying in Class C airspace which requires you to maintain two-way radio communication (§91.130(c), check it out in Ron Beyers answer).

This is another one of those questions where I would be interested to know exactly how many uncontrolled fields underly Class C, and which don't have an RCO.

More than likely, the best answer is going to be talk to other people at the airport and find out what they do.  There is probably an RCO or the telephone number to the tower is posted in the FBO.  Of course, if you don't do that, you are 100% allowed to take off and establish contact "as soon as practicable", and the tower is also allowed to tell you to exit the class charlie to the south and remain clear.  In the real world, I'm betting that there is a charted exclusion for the satellite airport in question.

This is spelled out clearly in CFR 91.130:

(c) Communications. Each person operating an aircraft in Class C airspace must meet the following two-way radio communications requirements:
(1) Arrival or through flight. Each person must establish two-way radio communications with the ATC facility (including foreign ATC in the case of foreign airspace designated in the United States) providing air traffic services prior to entering that airspace and thereafter maintain those communications while within that airspace.
(2) Departing flight. Each person—
      (i) From the primary airport or satellite airport with an operating control tower must establish and maintain two-way radio communications with the control tower, and thereafter as instructed by ATC while operating in the Class C airspace area; or
      (ii) From a satellite airport without an operating control tower, must establish and maintain two-way radio communications with the ATC  facility having jurisdiction over the Class C airspace area as soon as practicable after departing.

You can try contacting them while on the ground, but if you can't raise them, you have to do it as soon as you can after take-off.


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Do some light sport aircraft have clipped wings to lower their cruise speed?
Light sport aircraft may have a max airspeed in level flight at MCP of 120kts. I have heard that there are some light sport aircraft that have had their wings clipped in order to have a lower cruise speed.
But why do they clip the wings to reduce the cruise speed? Wouldn't that require a higher speed as less lift is produced at the same speed with less wing area?
Also, thinking about gliders having a big wing span and fighter jets having a short wing span I can not see why you would clip the wings? What am I missing?

Gliders have very long wings, but they also have a very narrow chord. Fighters have very short wings, but a (relatively) very wide chord.

@rbp: read all about it

You must have heard wrong. Clipping wings is done to reach higher maximum speed.
You mention light sport aircraft, so I assume they have propellers driven by piston engines. This means their power output is constant over speed and the thrust is inverse with speed. To calculate their optimum cruise speed, you can use the venerable Breguet equation which shows that the aircraft has to fly at its optimum L/D for maximum range. At this speed, induced drag equals zero-lift drag, and span helps to cut down the former. Clipping the wing will shift the optimum cruise speed up, not down.
With clipped wings the aircraft has less surface area, which will reduce zero-lift drag. This is the drag component which determines maximum speed. Induced drag at a given speed will go up, but it is insignificant at maximum speed, so the aircraft will be able to fly slightly faster.

Alright that's what I was thinking, just heard a experienced flight instructor say that so I didn't want to tell him that I thought he was wrong.

@Maverick283: I respect flight instructors when it comes to practical knowledge, but I have long stopped believing them in questions of aircraft design. There is too much nonsense circulating, and most of it is easily disproven. Still, people keep repeating it.

It can also improve roll response and roll rate

@JonStory: You are right, and much more can be said about the effects of clipped wings. This time I wanted to keep the answer short ;-)

Note that the Hughes H1 racer had two sets of wings: a short set for the high speed record flights, and a longer set for the transcontinental speed effort. Given that the power plant wasn't changed, the longer wings allowed for a more efficient high speed cruise, meaning fewer fuel stops, while the short wings were for maximum top speed in a short burst, where fuel consumption wasn't an issue (unless you let it run out of fuel during the attempt, as Hughes did).

Keep in mind that less wing area means a higher stalling speed, which translates to more runway needed for TO & landing. Climb performance and max. altitude may also be reduced noticeably.

Clipping the wings may help a plane meeting the max gross limit of the light sports, but there are also max stall and max cruise speed limitations, which would be adversely affected by clipping the wings, as Peter explains in his answer. A plane that already has very low stall and cruise speeds, but it pushing the max gross weight might be made to fit the light sport limits by clipping the wings. A J-3 perhaps?

Isn't official max limit simply imposed in the manual? Of course, it has a physical source, but just having larger wings and thus being able to lift more doesn't always mean it is allowed to lift more (and have greater MTOW); there may be many other constraints. Or are you talking about cutting the wings as a measure to save dry weight?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What fraction of fuel is consumed just for keeping a plane in the air?
In a typical airliner, what fraction of fuel/power/energy is used for providing lift and overcoming lift-related drag
Do the density of the air and the speed through it influence this ratio or do they affect lift and drag equally? If they matter, assume typical cruise altitude and speed.

To illustrate, if you took such a plane and:

attached its engines directly to the fuselage
removed all wings and ailerons (essentially only keeping the fuselage)
let the fuselage run along the flightpath on frictionless rails

how much more efficient would it be? Nothing else changes, so it's still in level "flight" at the same altitude and speed.
Optionally, in what ballpark is the amount of extra fuel consumed to initially climb to that altitude, minus any amount saved when descending to land?
Is it significant enough that, for instance, the length of a flight significantly affects the average fuel consumed per kilometer traveled horizontally?

I'd be happy with any answer that fits the description in the first sentence, but if you want details, I'll just pick:

Boeing 737 MAX 7
Altitude FL390
Speed 0.789 Mach
Payload 20'000 kg

Comments are not for extended discussion; this conversation has been moved to chat.

EDIT: I've found the polar for the B747 :-) Relevant calculation appended at the end of the answer.

Very little since aircraft are extremely optimised.
Engine's fuel consumption is proportional to the generated thrust, so the question becomes how much thrust is needed when the lift is null. At least for level flight, thrust equals drag and therefore this can be rephrased as: how much is the drag coefficient when the lift coefficient is zero, i.e. how much is $C_D$ @ $C_{L=0}$.
So: FC $\propto$ T = D $\rightarrow$ $C_D$ @ $C_{L=0}$ (FC = fuel consumption)
To answer this, the drag polar of the aircraft must be known. Online i could find only the following polars from this source:

Let's take for example the plot with diamonds for the F-16, since I like this plane :-)
The $C_D$ @ $C_{L=0}$ is some 0.022 while at "load factor=1.0", i.e at level flight, it is 0.025. This corresponds to a 12% reduction in drag (and thrust) when the lift is null. This 12% represents the share of drag which is correlated with lift.
How does this reduction of drag/thrust affect the fuel consumption? For the Mach number and the height given in the polar (M=0.9 at 30,000ft), the fuel consumption of a typical afterburning turbofan is more or less linearly proportional to the thrust and therefore our flying-with-no-lift F-16 would need some 12% less fuel.
So, the fraction of fuel consumed for keeping a plane in the air is just this 12% of the total, very little indeed!

Here is the polar of a B747-100 taken from the legendary Dr. Jan Roskam, Airplane Design Part VI, DARcorporation:

So, following the same procedure as before,
$C_D$ @ $C_{L=0}$ is 0.019 at Mach 0.86. For the cruise condition I used some halfway values i.e. height of 10km, Mach as before and mass of 250,000kg. This gives a $C_L=0.38$ and a relevant $C_D=0.023$ which corresponds to a -17% in drag i.e. thrust.
For a typical high-bypass turbofan at that height and Mach, the fuel consumption doesn't change linearly like for the afterburning turbofan of the F-16 but a bit less than linearly. So the 17% reduction in drag/thrust translates into a 14% reduction in fuel consumption.
Then again: very little indeed.

Thanks, I have to say that's a lot less than I expected. Of course that data is with (now redundant) wings still attached, but I presume removing those wouldn't put the results in another ballpark.

I'm curious though, how did they find those numbers in your source? Specifically, how/when does a real plane, with the wings attached and moving through the air, have a lift coefficient of 0? If I understand it correctly, it can be increased by using e.g. flaps. But the wings are fixed. I guess by compensating the lift with ailerons, or transiently by flying with the roof tilted towards the incoming airstream a bit, using momentum. But surely those greatly affect the drag as well, and at different rates too? I guess my understanding isn't quite complete yet.

@BartvanHeukelom: Note that this is at Mach 0.9.  Going that fast creates a lot of drag.  If you want to save fuel, go a lot slower, especially if you're getting lift for free.  (Like a train, where the wheels keep it off the ground with minimal rolling resistance regardless of speed.)  You're talking about a plane which has these big bulky wings sticking out into the air, creating drag for no benefit (because you're ignoring lift requirements), not a bullet train, so you want to go slower to reduce total energy lost to drag over the distance you have to travel.

It’s not the drag at Cl=0.  There is no Cl. So you can choose the alpha that has minimum drag, not the alpha that gives Cl=0.

@Jim: since in the question the lift is created for free then yes. Anyway the 2 values are really close (at least in the polar I posted) and I wanted to stay "real" ;-)

@BartvanHeukelom: if we remove the wing as well then the question would become if it's more streamlined a pressurised tank or a bullet train, not really an aviation question :-D  Anyway for a fighter like the F-16 there's not really a fuselage and a wing but the whole surface is optimised to create more lift and less drag. It would be nice to have the polar of a jetliner

@BartvanHeukelom: polars for the whole airplane are a mix of wind tunnel tests, CFD and real flight data

@BartvanHeukelom: anyway very interesting question

Thanks for going the extra mile and digging up the 747 data! Note that I marked your answer as accepted after you did, but would have done so without the addition as well. In fact, I only noticed it after I accepted it.
"So the 17% reduction in drag/thrust translates into a 14% reduction in fuel consumption." - not relevant to this question, but out of curiosity: do you know where the other 3% goes, so to speak?

@BartvanHeukelom: I'm glad I could help :-)  The lost 3% basically goes into inefficiencies; this answer should clarify that.  P.s.: one cannot choose or upvote his/her/its own answer or question. I don't know who did it

@sophit I was unclear. By "after you did" I meant "after you added the 747 bit"

@BartvanHeukelom: ah ok, sorry :-D

Airplanes not only have to hold themselves up, they also have to overcome drag from speed.  But to give you some idea, ton miles per gallon is used in the transportation industry.
Trains are at least twice as efficient than tractor-trailers, who are around 4x more efficient than a 747, which (because it can fly high) is around 2x better than the best 4 engine piston prop era aircraft.
The selling point of aircraft is speed. Ground transportation is slower, but more efficient for one trip.
The aircraft, with its speed, can make 10 trips for each one a train or truck (forget about a barge unless it's your yearly harvest) can make.
In the end, the analysis goes to the bottom line profit for each trip × the number of trips made per given amount of time.  A FedEx jet stuffed with overnight mail will probably be a huge money maker.

I would also add container ships, which are the cheapest way to move goods around the world

Oh yeah for sure, land or water based transport must be many times more efficient using that measure, or it wouldn't be that popular. But that efficiency is affected by many variables, even when ignoring things like the cost of requisite infrastructure. (1)

I'm trying to break the comparison down into the separate components a bit. I figured speed would be a very significant factor considering the exponential losses of drag, but that's suddenly irrelevant if you compare the plane against e.g. some kind of future maglev train thing that travels just as fast, through similar air (by whatever means). So I wondered what the next big difference would be there, and guessed maintaining lift would be a good candidate. (2)

@BartvanHeukelom: don't forget density: jetliners normally fly where density is more or less ⅓ the one at sea level. So in your hypothetical aircraft generating lift for free, drag reduces of the same factor


text:
Does this Hawaiian Airlines descent look routine?
I was on flight HA 30 from Kahului to Seattle on Dec 19 2021 (Dec 20 Zulu time).
FlightAware track log
FlightRadar24 track log
The descent looked anything but familiar.
As you can see it reached over 5,400 fpm at times.
I definitely felt it.
After descending from FL410 to 10k MSL, the plane got back to 16k MSL before landing.
I have looked at the flightradar replay and it looks like the plane is heading to Portland, it follows the very same path as ASA834 and HAL26 (both Honolulu-Portland flights) in front of it. Could it be that it mistakenly started to descent towards Portland ?
HAL 26

ASA 834

HAL30 at 10k MSL

Full descent path
.
All other flights coming from the Hawaii island towards Seattle that evening crossed into mainland way further north: HAL 22, DAL 489, ASA 942.
Do you find all this strange, could there have been an emergency or misunderstanding ?
Additional details, in response to questions asked:

We did have 2 stints of small hail hitting the airplane while descending, don't recall exactly when.
At the very start of the descent, the seatbelt sign was not on. A couple minutes later it went on and the plane immediately started descending visibly faster.

Update
Got a response from Hawaiian Airlines Consumer Affairs Office

We've contacted our System Operation Control Center (SOCC) and they've stated that the weather in the area of Portland Aiport was poor and the dispatcher routed the plane to not encounter the poor weather and turbulence in the higher altitude around Portland.

There must have been a legit reason for the odd descent sequence, and hearing this from someone in the know would be interesting. Voting to keep open.

@Koyovis but the question asks "does x look routine" not for the explanation of someone in the know

Of note: FlightAware (which is usually accurate in this regard but not infallible) shows the route in question as CLUTS R465 CORTT ADTIL 3800N/13200W 4000N/13000W HEMLO ONP J136 BTG HAWKZ7 (Skyvector link). Other flights before and after the flight in question use APACK R463 AUNTI A332 HELOP A332 HEKAB FOZZI SEDAR LATAY HAWKZ7 (Skyvector link).

Why we don’t ask the airline or pilots? Best opportunity is after the flight but also afterwards. It is not about blaming but curiosity.


text:
What is the buzzing noise generated by the engines in this 777 takeoff video?
I've flown a few times in my life, both in economy and executive class. Every time I fly in executive, during takeoff, there is a different noise from the engines. In economy class (behind the engines) the wind sound dominates, but in the executive class you hear the noise in the video below until you're high in the skies. Why is that?

Behind the engines is also behind the wing.  The wing disrupts airflow (the wind noise).  Behind the engines much of the engine noise is muffled by the wing and engine cowl.

Related: Why do most A320s make that distinct scream on takeoff?

That buzzing sound is from the tips of the fan blades breaking the sound barrier.  When you're sitting farther back in the airplane, that sound is overpowered by the sound of the exhaust, which is the roaring part of a jet engine's sound.
So in general, sit forward of the wing, you'll hear the fan sound (the buzz), sit aft of the wing and you'll hear the exhaust sound (roar).  It'll vary from airplane to airplane of course, and when thrust is reduced and the fan speed slows, the buzz will subside.

Do you know why the engine suddenly changes sound at 1:45 ?

Thrust was reduced there.

I notice that sound every time I fly, and I always thought it was either the thrust that was reduced or it was the air pressure/clouds that changed the sound of the engine. Thanks Ralgha!


text:
How can the Airbus A319 accommodate 156 passengers with overwing emergency exits?
This official image from the Aircraft characteristics airport and maintenance planning document for the A319 advertises the A319 as accommodating 156 passengers in a high density layout, this is also the figure in other Airbus documentation.

What I would like to know is how emergency exists is accommodated for? In the image it shows a row of seats at the exits, but I was under the impression that there is not allowed to be seats in line with the emergency exits? Can this variant thus only accommodate 150 passengers (156 - 2*3)?
My question is also valid for the A320.
Then I would also like to know, is there a policy for the amount of lavatories on board the aircraft, or is that simply a choice made by the airline? Are there for example airlines that have only one or even none lavatories on board models only used for short hauls?
Thank you for any help.

that's a double exit over the wings with seats in between

You can just barely see that those two rows have a bit more legroom for the exit.

So is this permissible in FAA and EASA regulations? I'm sure there is a technical regulation governing this? I doubt if 'a little more legroom' will be a sufficient specification (-:

I don't think I've ever been on a jet airliner that didn't have seats in line with the over-wing emergency exits.

Same seats at dual over-wing exits on United B737

It seems like lavatories are a separate issue, and should be split into a separate question

Yes, airlines can use a 156 seat layout in the A319. See Allegiant's layout.
The FAA type certification for the A320 series can be found on the site here, under TCDS A28NM. For the A319:

Maximum Passengers:
  145
  See note 11 and 18
  ...
  Note 11:
  If modification 32208 is embodied in production on A319 airplanes, the aircraft is eligible for maximum capacity of 160 passengers. This modification consists in structural and system provision for the installation of second pair of overwing emergency exits.

The certification basis section g states that the design complies with the following regulations:

25.811(e)(3) Type III emergency exit marking.
25.813(c) for emergency exit access

The regulation from section 25.813:

§25.813   Emergency exit access.
(c) The following must be provided for each Type III or Type IV exit—(1) There must be access from the nearest aisle to each exit. In addition, for each Type III exit in an airplane that has a passenger seating configuration of 60 or more—
    (i) Except as provided in paragraph (c)(1)(ii), the access must be provided by an unobstructed passageway that is at least 10 inches in width for interior arrangements in which the adjacent seat rows on the exit side of the aisle contain no more than two seats, or 20 inches in width for interior arrangements in which those rows contain three seats.

The A320 series has three seats on each side, so the applicable distance is a 20 inch wide passage. The referenced paragraph (c)(1)(ii) is also interesting:

(ii) In lieu of one 10- or 20-inch passageway, there may be two passageways, between seat rows only, that must be at least 6 inches in width and lead to an unobstructed space adjacent to each exit.

This appears to be one layout that Air Canada uses.

There doesn't appear to be any regulation requiring a certain number of lavatories. Aircraft sometimes fly with only one working. Evidently no US agency requires even at least one.
However, the DOT does appear to have a rule created by the Air Carrier Access Act that says:

New twin-aisle aircraft must have accessible lavatories.

I read that as requiring the aircraft to have at least one, possibly two, lavatories, and that they must be accessible.

"New aircraft" requirements apply to U.S. airlines with respect to planes ordered after April 5, 1990 or delivered after April 5, 1992. In general they apply to foreign carriers with respect to aircraft ordered after May 13, 2009 or delivered after May 13, 2010.

Of course this wouldn't apply to the A320 series, as it is a single-aisle design. Although some twin-aisle aircraft do operate on shorter routes, they are probably less likely to go the route of ">200 people, no toilets."


text:
Are general relationships between MTOW and OEW valid for novel configurations?
I am a student in mechanical engineering with a passion for aerospace engineering, especially airplane design.
However, coming from a more mechanical than pure aeronautical background, I feel that even now, doing a masters in what is essentially conceptual airplane design, I am lacking a certain foundation in the methodology of airplane design.
To remedy this I have recently obtained the very well-known books on airplane design by Roskam and Raymer and have started working through these books.
My question is: In the modern age of airplane design, where we are starting to get interested in more novel and radical configurations, are the approaches presented by Roskam and Raymer (and I am sure there are others authors) still valid?
To summarise this design methodology I am talking about:

Determine payload weight
Guess aircraft maximum takeoff weight
Estimate fuel fractions for different mission segments from historical correlations
Calculate tentative empty weight
Determine allowable empty weight by taking advantage of the fact that, historically, there is a logarithmically linear relationship between maximum takeoff weight and empty weight
Compare tentative empty weight and allowable empty weight, if the values are too different, guess a new value for maximum takeoff weight and continue this iterative process until convergence

This design procedure uses historical correlations in the fuel fraction determination and through the historical trend in relationships between MTOW and OEW.
Is this approach still valid for novel configurations?
Most importantly, is it known whether this logarithmically linear relationship between MTOW and OEW independent on configuration?

I suppose the methodology would still work, but these days the constraining factors on aircraft designs are often not "aerospace engineering" issues but human or infrastructure ones: The 787 could have been faster, but it needs to meet noise regulations (people don't like loud planes flying overhead). The A380 could have been bigger, but its size already causes trouble operating at some airports. The improved design technologies (e.g. simulation) that Peter mentioned help contribute to working more of that out on paper before building the flying prototype.

Many studies and presentations you will encounter about radically different configurations with much higher performance are the aviation equivalent of car commercials. Someone tries to get attention in order to get published, or she/he needs to convince investors to fork over their money.
Trust me, the aircraft designers of the last century were no idiots. They had the luck of living in a time when it took only a few years for aircraft designs to become obsolete, so they could go through the whole design process many times and apply what they had learned before.
Someone graduating from university today will be lucky if she/he gets the chance to go through every step of a major design process even once. I myself have brought three aircraft into the air and consider myself very lucky. I had colleagues who retired without ever having seen one of their many projects take off (literally).
The consequence of this is a very mature industry which has found an optimum configuration for most purposes already and is only tinkering with the details. Add to this an ever increasing thicket of regulations which have grown on the experience from existing designs. Sure, the progress in computer control is impressive, and engines get better all the time. Manufacturing precision is ever increasing, and materials still become better and more consistent. But overall, we still build aircraft like we did a generation ago, and I do not expect this to change radically in the future.
What impacts the design process most is certainly the incredible precision of simulation software, which allows to see details which remained hidden in a wind tunnel or a strength test. Things which had to be done consecutively can now be done in parallel, and the precision of knowledge at an early stage of a design is much higher than before. But at the same time aircraft become ever more complex, so the advantage of better simulation is eaten up by the increase in complexity. Add to that the fact that mostly beancounters will have the last word, where before engineers could determine how the work was done, and you will understand that the level of preparatory work is shockingly inadequate in modern aircraft companies. Many of the avoidable early mistakes will require late, expensive fixes (but by that time the stingy beancounter has left the company with a huge bonus for the money he bragged to have saved).
The design process is still the same, and where before the experience of a few people helped to find an overall optimum, large, integrated teams will together determine this optimum. The inexperience of the participants will only show up in delays, when the iterative design cycle needs to be repeated more often than anticipated, but the result will be of high quality - really bad designs are a thing of the past. But cutting corners will still have consequences, so small annoyances will still plague new designs. Consequently, most of the engineering work will involve ironing out these annoyances or retrofitting new systems to existing airframes.
To answer your specific question: Yes, the relation between OEW and MTOW will still hold in the future, and radically new configurations will show an advantage only on paper, before they have been thoroughly designed and test-flown. Better materials and methodologies will help to improve the ratio between OEW and MTOW, but this advantage risks to be eaten up by the desire to add bells and whistles everywhere.

@Jonny: It is good practice around here to accept an answer only after two or three days, so you have a choice of good answers and don't discourage others to add their view.

Noted. Also after asking my question I realised that I might evoke some opinions rather than clear answers, which is also a bit against the grain of SE. But I think your answer was specific and technical enough (-:

What does it mean OEW ?

@d.pensopositivo: Operating empty weight. This is structure + systems + pilots + equipment, but no fuel, lubrication and payload. Also called OWE (operating weight empty).

To add to @PeterKämpf's answer: the basic design methodology will be unchanged so long as the laws of physics remain unchanged. But the method of application of the methodology is changing very fast, and will continue to do so.
As a mech eng student, the OP is unlikely to have any real-world experience of the engineering environment in which Roskam's 8-volume set of books were first published, in 1985-1990 (according to Wikipedia). That is 25 to 30 years ago. As just one example, the fastest computer in the world back then (the Cray-2) had a clock speed slower than any modern PC except for cheap laptops or tablets, and less RAM than an top-of-the range modern cellphone. And with a price tag of around $20m, that computing resource would be shared between hundreds or perhaps thousands of users, not sitting on an engineer's desk for his or her exclusive personal use.
Today you don't need to "estimate fuel fractions for different mission segments from historical correlations". You can look at the accurate real-time data being collected automatically by every modern aircraft that your company has built, and look at it for negligible marginal cost. You don't need to approximate from a "historical log-lin correlation" - again, you can look at the actual data, in as much detail as you need.
This accelerating rate of change presents educators with some serious challenges. The university engineering departments don't have access to all that real-time data, and even if they did they are (correctly, IMO) trying to teach some general principles, not the specifics of what Company X does this week. But the danger is that the students work through the case studies in an excellent collection of 25-year-old books, and without any other input, they get the wrong idea that those case studies still represent the current state of the art.

Thank you, very valuable insight. I suppose that as someone wanting to work in the industry it wil nevertheless be valuable to get grips with the methodologies, whether their implementation is the same as in the days of Roskam or not. I know that both Roskam and Raymer currently market design codes based on the methodologies and methods presented in their books.

So they must still be valid (the methods as well as the implementation). I think one should realize that not all aircraft manufacturers are the size of Boeing, Airbus, Embraer or Bomabrdier etc and that many of the smaller companies still us these methods and correlations as is. Of course this is just my thought, not even an opinion.

The term “historical data” does not mean the data is old, it simply means data collected from previous actual use as opposed to data calculated or measured in a specially constructed experiment. Of course you'll use the most recent data available. Also detail is not always what you need; it just gives you more accurate estimate of the “historical log-lin correlation”.

Add to the improvement in computing power the improvement in algorithms (which, by the way, has been bigger than that in computing power over the last 30 years in a number of applications), and you can see what the change has been. But it still needs an educated mind to understand the results, no matter how pretty the colorful output will be.

What does it mean IMO and OP ?

IMO = In My Opinion.  OP = Original Post.

@MikeBrass: OP can also mean Original Poster.


text:
What is a biplane propeller and how efficient is it?
Lazair ultralight series II used biplane props to absorb more power from the engine....so my question is what really is a biplane propeller and are they efficient in comparison to a four bladed prop?

It is essentially two propellers stack on top of each other. You can see it in this picture of a Lazair.

As for why they were chosen for the Lazair,

Many have asked why Ultraflight opted for double propellers stacked in
  a biplane fashion. This is a very unique setup and draws lots of
  attention. Apparently when Ultraflight changed over to the Rotax 185
  they tried several different propellers. The problem was the 185’s had
  a woodruff key slot in the crankshaft and would snap the crank
  whenever the engine coughed or kicked back on start up. This was quite
  a serious problem. After much deliberation it was determined that the
  rotational inertia of the propellers was too much for the crank to
  handle.
Dale could not locate a two bladed prop light enough to fit the need.
  However, since they had been producing a lot of the early lazairs with
  Pioneer engines fitted with a single plastic prop, he had a ton left
  over that fit the pioneers.  He found that these could be “doubled up”
  and help to produce the extra thrust that the Rotax 185 was capable of
  producing over the 5.5 hp Pioneer engines.

It appears to be a quick way to gain more thrust from the props without increasing prop mass too much.

… and I'd wager a guess that they are stacked in this fashion to avoid having to drill some more holes in the central disks when it worked sufficiently without—mounting them across like a four-bladed propeller would be more efficient and weigh exactly the same.

@JanHudec you are correct. The linked article actually discusses that. Longer through bolts were used on the Lazair which ultimately lead to some serious fatigue issues and eventually they started throwing props.

I'm stunned that the close spacing does not produce unacceptable aerodynamic interference!

Depending on the pitch and blade airfoil, a biplane propeller as used on the Lazair might be more or less efficient than a conventional four-blade propeller the same diameter.  The interference between blades causes additional drag (which absorbs power without creating additional thrust), but the turbulence from each blade affects the following one in a conventional four-blade design, with similar effects, especially in a flatter-pitch design or at low airspeed.
The potential advantage of the biplane propeller is that, presuming the hubs have six or eight torque transmission bolts (or even four, though that arrangement is less versatile), the blades may be arranged with the forward propeller directly ahead of the aft (true biplane propeller), or one or two holes leading or trailing in rotation.  Without wind tunnel testing, it's impossible to be certain, but I'd expect an arrangement with the forward propeller trailing the aft by a fraction of a revolution to be most free of interference effects and hence most efficient, quite possibly more efficient than a four blade design with the same diameter, pitch, and blade area.

A biplane propeller is one prop stacked above another, similar to a biplane wing.
This looks to be very inefficient. The gap between stacked blades looks like about one chord at the tip and a quarter chord at the root. This should cause problems with the high pressure under the upper blade interfering with the low pressure above the lower blade. At minimum the blades should be spaced farther apart.
Here is an analysis of biplane wing efficiency relative to separation as a percent of chord. Efficiency is 10.5% higher when the gap/chord is 1.5 as compared to 0.75. It doesn't go to 0.25 but it seems that would be much worse as the 1.5-1.25 gap drop in lift/drag is 3% and the 1-0.75 drop is 4.2% at 4%. This configuration is likely losing 6%-18% compared to a gap at 1.5 chord. The .075 gap/chord entry in the table looks like a typo.
The reference shows biplane efficiency is lower than a similar monoplane, though not completely clear if wing area, span or chord are held constant for that comparison. It also states that positive stagger (upper wing ahead of lower) by 0.4 chord gives 5% better efficiency, so rotating the upper prop 0.4 chord would be beneficial compared to the vertical stacking shown.

A drop in absolute efficiency may not be a bad thing in this specific case - clearly the 'unusual' choice of prop setup is intended to match a pre-chosen engine , with available props, without modifying either engine or prop. By which I mean, a pair of props with low efficiency may give better "system" efficiency than a more traditional choice.


text:
Is flying in ground effect dangerous?
The stall angle  of a wing in ground effect reduces in comparison to the stall angle of a wing flying outside of ground effect. Doesn't this make soft field takeoff dangerous?

One typically doesn't increase AoA to gain speed.

Not if you are an ekranoplan

@Dave: Nor if you occasionally fly around dry lake beds and such :-)

The decrease in stall AOA is more than offset by a decrease in the AOA required to achieve the same total lift and the result is indicated stall speed actually drops a little bit.  And pilots generally don't fly AOA itself as a parameter, but pitch and speed.
But the main thing is the power required to fly is less in ground effect so there is more power available to accelerate.  Normally when taking off you rotate and start right up because even though you are initially below best climb performance speed, there is ample reserve and you accelerate to it with no problem. Problem is in marginal conditions, where the airplane won't climb at all unless it's at its most efficient climb speed, you need to accelerate somehow without trying to climb, and you can't just roll along the ground faster, so you take advantage of the power bonus you get in ground effect.
Ground effect is much stronger the higher the aspect ratio.  How strong?  Watch this glider's L/D go through the roof:

That video just totally rules!


text:
Bed sheet as wing skin?
https://en.m.wikipedia.org/wiki/Aircraft_fabric_covering
This article states several early aircrafts that successfully used cotton as wing skin including Wright flyer.  "Colditz Cock" glider had its wingskin made from bedsheets plus this guys

 used bedsheets to make their wingskin and the plane kinda took for like 2 seconds.
Can a  good,strong bedsheet successfully replace tedlar in an airplane without any problem?

Depends on what you mean by "replace".  After all, there were fabric-covered aircraft for decades before there were synthetics like mylar &c.  Though I would look at a fabric optimized for strength rather than sleeping comfort, myself :-)

Trying to cut costs or something?

The earliest covering systems were actually bedsheets... well not literally bedsheets, but the fabric was linen.  The Wrights used muslin, a tightly woven cotton used for womens' unmentionables. Grade A cotton came later after linen, then polyester after WW2.
I'd say the cheapest alternative to Tedlar might be to use commercial grade polyester fabric of a suitable weight that can be heat shrunk and use premium exterior latex house paint as the filler system. You'd have to use polyester instead of cotton or linen because on those fabrics it's the dope coating that shrinks the fabric to make it taught and I don't think latex paint will have the same shrinking effect.
Instead of aluminum powder added to the dope for UV protection as is done for normal airplanes, for latex you add carbon black to the base coats.
Anyway, I'm not making this up.  Latex paint has been used on homebuilt aircraft with success http://www.flysquirrel.net/piets/paint/paint.html.  I've seen that Peitenpol in the article at Oshkosh and it looked pretty good.  The article notes that the latex finishing system was actually lighter than an actual aircraft finishing system like Polyfibre.

Absolutely you can use a fabric for covering and airframe.  I used to build large scale R/C aircraft much like they used to build real world aircraft. I used fine silk to cover the wings, elevator, tail, and fuselage. I used epoxy to adhere it to the wooden airframe and then later apply some kind of lacquer, which serves to shrink and pull tight the silk and also makes it airtight. This can then be painted or not.  Silk is a very strong lightweight material. The performance is actually probably better than many modern materials, but probably would not hold up to the elements as well over time.

Welcome to aviation.SE!

Fabric coverings need to be strong enough to handle the load to which they are subjected. As an example:

Piper PA–18s, for example, emerged from the factory wearing Grade A cotton fabric tested to withstand a minimum of 80 pounds per square inch.

This super cub cruised at 130mph with a wing loading of about 10lb/sqft.
Fabric wings, particularly cotton, are often also "doped", meaning they are painted with a material that shrinks, preserves and makes the fabric less permeable to air.

Classic and antique aircraft got coats of plasticized lacquer called dope, a flammable material that served to tighten the fabric and provide some protection from the elements; paint was applied over the dope.

Some ultralights with UV resistant nylon coverings do not take this extra step. If your aircraft is only to be used a couple of times, preservation is less important.


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Is this type of wing design structurally sound?

Is the simple wings
frame(spar,ribs)of this maxair hummer structurally sound? I am asking because I'm considering using thesame wing design on my homebuilt ultralight.

The wing has to withstand forces without breaking and at the same time have little weight. A very simple construction will be heavier than an optimized, more complicated design. Your choice.

This looks like a standard fabric-covered construction. That's how all aircraft were built in the first, say, 20 years (except the ribs were full, which is the only way to make them from wood and would probably be better when using metal too). Simple, but not extremely so (constructions like Rogallo wing are simpler still).

The simple answer is yes. There are many examples of working tube and fabric aircraft, both ultralights and larger designs, that make use of wing designs similar to your example.
The more complex answer is that it depends on a lot of things:

Did the designer anticipate flight loads correctly?
Did the designer model flight loads correctly when validating the wing design?
Do the materials used to construct the wing have physical properties close enough to the model used during design?
Was the aircraft built correctly according to the design?
Was the aircraft damaged such that the structure was compromised?

I don't want to discourage you from designing and building your own aircraft--one of the most exciting things about ultralights is how accessible they are to people with little or no formal training. Before you build and fly though, you should either learn how to determine the safety of a wing design using analytical methods, or stick to copying proven designs exactly.


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What is the average g force experienced on takeoff for an ultralight?
How much g-force is made on an ultralight during take off if its upward velocity of 2m/s?And what's the average g force experienced during takeoff on an ultralight?

about 1.2g? (9.8+2)/9.8 = ~1.2 ?

nope, g=9,8 (m/s²) is acceleration while 2 m/s is velocity. totally different things and cannot be summed together.

Assuming wings level takeoff (no 60 degree banked turns during takeoff), Average (over the course of entire takeoff), is probably very, very close to 1 g. (starts at 1 G, ends up at 1 G, so there must be periods where actual G-Loading is slightly above 1 G and periods where it is slightly below).   But based on 2m/sec, (2 m/sec over a 5-10 sec takeoff period is what in acceleration? 0.2 to 0.4 m/sec/sec?) Maximum G-Load however, is different story.

... and @szulat, acceleration is simply rate of change of velocity, so user3528438 is correct (although he makes incorrect assumption that takeoff period is only one second.)

@CharlesBretana exactly, it's the rate of change and in this case the rate was not specified so the g force can't be calculated. if the user35 said the assumed time is 1 s then it would be correct. adding acceleration and velocity numbers together and hoping it somehow lead to meaningful result without understanding what actually is being calculated is incorrect.

Well, all you have to do is make an assumption about how long it takes to take off in an ultra light. I looked at some YouTube videos, and Ultra-Light takeoffs only take between 5 and 10 seconds to get airborne - add another five or ten seconds or so to establish a vertical velocity of 2m/s, and you can assume (within an order of magnitude at least) about 10-20 seconds.

not necessarily, because the vertical acceleration is obviously zero during the whole ground run so those initial 5-10 seconds do not count at all. then, at the moment of rotation the vertical velocity is quickly increased, which sounds like a short but significant g-force impulse. highly variable acceleration is not something that can be reasonably described by a single average value. i will try to measure these things with a smartphone accelerometers on the next opportunity...

You're asking about takeoff, and you illustrate (or whatever one should call it) your question with a picture of someone in a ~90° bank at altitude?

Funny enough.....am just seeing that....

I fly part 103, specifically a Hy-Tek Hurricane. I also have a G meter on board. It is a Dynon D2, which features an artificial horizon (attitude indicator) and also a G meter. The average G's on takeoff is pretty small, usually around 1.1 to 1.2 Gs. I can also tell you it is impossible to pull anything over 2 G's during takeoff. I actually performed this test in the air at 2000ft, straight and level at full throttle (simulated takeoff from 2000ft?) at about 55mph, then pulled the stick back all the way in about 1/4 of a second until I was nearly vertical. The highest recorded G reading during this maneuver was 1.8 Gs. Of course all ultralight aircraft are different. I would NOT recommend performing this test from an actual take-off from the surface. Also you should consider that by performing the test at 2000ft, I was not in any kind of ground effect, so you would have to factor in that little bit of extra lift cushion you would get from an actual takeoff from the surface, but I would guess the extra lift component is probably small, so that is why I estimated 2 Gs max.

55 mph is, however, way too fast for take off, isn't it? At take-off speed, you couldn't pull that much, because you would stall if you did.

Exactly Jan. Since normal take-off speed in my aircraft is lower than 55mph, I would never reach 1.8 Gs on takeoff. If I tried yanking the stick back on actual takeoff, I would possibly stall the wing immediately. I don't know for sure, but I am not going to try.

Given that lift goes as the square of speed, at 35 mph, your maximum G would be around half what you pulled -- which is less than 1 G (i.e. you couldn't be off the ground yet).  At the moment of takeoff, there's precious little extra lift above the takeoff weight (including pilot and fuel) -- so G figures in the 1.1 to 1.2 range are very reasonable.

The answer depends on how quickly was this 2 m/s velocity reached.

Could be 2 g if it happened within 0,2 s (9,81m/s² + 2m/s / 0,2s)
Could be 1.1 g if we were accelerating for 2 seconds to get to 2 m/s (9,81m/s² + 2m/s / 2s)

Example accelerometer recordings taken during the Cessna 152 takeoff (not exactly ultralight, but also a small plane). The first noticeable spike was about 1.2g
(recorded on iphone, 30 measurements/second)

And 0.2 s is absurdly short—if you handle the controls gently and smoothly as you are taught, you will definitely not rotate in 0.2 s.


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Is aerodynamic flutter speed dependent?
Is it possible for  control surface flutter  to occur at low Reynolds and low speed?
If yes,why don't Ultralights have counter weights or tail sting mass to immune the control surfaces against flutter just like how most G.A planes and aerobatic planes do?

Flutter requires 3 factors to come together: flexibility of the structure that a control surface is attached to; the natural frequency of said structure; and an undamped moving mass attached to the structure with an internal CofG offset from the hinge point of the mass such that inertia helps magnify motion of the mass.  If you have an ultralight with a really floppy, flexible wing that can twist easily, with a heavy, unbalanced aileron, and you excite motion in the aileron, which excites sympathetic motion in the wing itself, you can probably get it to flutter at 40 mph.
I say undamped moving mass because a lot of jets with hydraulically operated control surfaces don't use mass balancing but depend on the damping of the hydraulic actuators, and if that isn't sufficient, dedicated flutter dampers that work like motorcycle steering dampers connecting the surface to the structure.
Anyway, ultralights and other lighter airplanes are able to avoid the factors I mentioned because, in designing it to hold up the airplane in the first place, the wing/stabilizer is inherently stiff enough, and/or the oscillating mass attached is light enough, combined with the lack of potential energy in the system at the low speeds they operate at, so it's not a problem.
You see lots of light airplanes (the Aeronca Champ for example) that operate up to 100 mph + without balanced surfaces because the structural stiffness of the strut/wire braced wing/tail and the light control surfaces makes it impossible for the control surface to excite sympathetic motion in the wing/stabilizer, making balancing unnecessary.

Just a tip, @DavidTeahay - It's recommended to let your question sit for at least 24-hours before accepting an answer. Many people don't bother to read a question with an accepted answer, so you may miss out on additional information. NOTE: no implication whatsoever that this answer is wrong or even not very good (I don't have enough knowledge to say one way or the other), just that others may not come along now that there's a check mark.

Got it... @ FreeMan

FreeMan is absolutely right.


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What is the relation between drag and weight?
Given an aircraft type and fixing its wing configuration (flaps, slats) and fixing its speed and altitude and assuming it flies horizontally, what is the relationship between the aerodynamic drag and weight of the aircraft?
Of course this relationship depends on the aircraft itself and on many other things, but what is more or less the power-law?
Say we compare two A320s, the one heavier by a factor of 1.3. Of course the heavier one will fly with a higher angle of attack, because it needs more lift. This will create more drag and the engines will run with higher power to achieve the same speed. How much more drag will it be? If the behaviour is approximated as power-law, what would be the exponent? The cross section increases when we "pitch" the aircraft. Is the aerodynamics in the design process optimised for "half-loaded" aircrafts? Also, how does this translate to fuel consumption, say on 40000 feet with the usual speeds?

If you allow to make a few simplifications, the answer is easy:

Friction drag is not affected by the angle of attack change. That means no onset of flow separation on the heavier aircraft.
The wing's optimum L/D is reached with the same flap setting in both cases.
Lift changes linearly with angle of attack, so the lift coefficient $c_L$ can be expressed by the product of lift curve slope $c_{L\alpha}$ and angle of attack $\alpha$.
We neglect the changed lift contribution of engine thrust when the angle of attack is increased.

Now the drag $D$ can be expressed by this equation:
$$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\left(c_{D0}+\frac{(c_{L\alpha}\cdot\alpha)^2}{\pi\cdot AR\cdot\epsilon}\right)$$
Per the definition above the term $c_{D0}$ is constant, so the change in drag between the lighter aircraft (index 1) and the heavier aircraft (index 2) will be
$$\Delta D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\frac{c_{L\alpha}^2\cdot\left(\alpha_2^2-\alpha_1^2\right)}{\pi\cdot AR\cdot\epsilon}$$
In order to express this drag difference $\Delta D$ in terms of the aircraft's mass, write the lift coefficient $c_L$ as $\frac{2\cdot m\cdot g}{\rho\cdot v^2\cdot S}$:
$$\Delta D = \frac{g\cdot\left(m_2^2-m_1^2\right)}{\pi\cdot AR\cdot\epsilon}$$
The other symbols are:
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} S\:\:\:\:\:$ wing surface area
$\kern{5mm} c_{D0} \:$  zero-lift drag coefficient
$\kern{5mm} \pi \:\:\:\:\:$  3.14159$\dots$
$\kern{5mm} AR \:\:$  aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$  the wing's Oswald factor
$\kern{5mm} g \:\:\:\:\;$  gravitational acceleration
Now we can answer your questions:

How much more drag will it be?

The drag will increase with the square of the mass increase. The gradient of that increase depends on the span loading of the aircraft.

what would be the exponent?

2

Is the aerodynamics in the design process optimised for "half-loaded" aircraft?

No, always for the fully loaded aircraft, as lighter loads can be tolerated much better than higher loads. However, since fuel burn will cause a change in aircraft mass over time, the aerodynamics must work over a range of altitudes.

how does this translate to fuel consumption, say on 40000 feet with the usual speeds?

Drag is compensated by thrust, so you need more thrust to overcome the higher drag. Fuel consumption goes up linearly, but since cruise at 40.000 ft means that the aircraft is light and the engines are running at close to their maximum sustained thrust, a mass increase by 30% is impossible. For a practical result, the heavier aircraft would fly at the same angle of attack and speed but at a lower altitude where both the increased lift and the increased engine thrust can be provided by the higher air density.
If you need to calculate the trip fuel at different weights: We had a similar question before, so please follow the link for an explanation.

Why do you say, a mass increase by 30% is impossible? What is the mass ratio for commercial aircraft comparing start and end of cruise, when they operate on maximum range?

@AlessandroBrillante: No, all I say is that a mass increase of 30% is impossible for an airliner at its given cruising altitude.

It is hard to say what the ratio would be exactly.
Total drag consists of parasite drag and induced drag. The influence of angle of attack to parasite drag would depend largely on the aircraft design and is usually designed to be lowest at average flight condition weight and at cruising speed and altitude. However I think the effect of angle of attack to parasite drag will be minor at such low angles.
For the induced drag I can at least give a mathematical prediction.
Since everything in the lift equation but the angle of attack stays fixed in our example, angle of attack needs to be proportional to weight in horizontal flight.
We know that for small angles of attack the coefficient of lift is directly proportional to the angle of attack.
Induced drag is proportional to the square of the lift coefficient. Therefore the induced drag would also be proportional to the square of the angle of attack.
So 1.3 times the weight means 1.3 times the AoA means 1.3 times the Cl means 1.3^2 times the induced drag.
However at cruising speed induced drag usually accounts for half of the total drag.
So your formula would be something like:
Increase in drag= (50%)+ (50%)*1.3^2
Assuming engine thrust is proportional to its fuel consumption, the same formula could be used for fuel consumption
In conclusion weight plays an even larger role in the climb and during landing and takeoff where induced drag plays a major role. At high speeds weight is not as much of an issue.

"at cruising speed induced drag usually accounts for less than 10% of the total drag" are you really sure about that? We are talking about airliners here, and they fly close to their optimum range polar point where induced drag is half of total drag.

Thinking about it now, I think you are right it should be half of total drag. I should have thought about it more deeply instead of just putting in numbers of some article. Somehow didn't occure to me that I wouldn't even have had to look it up in the first place. Thanks for noticing!

You kind of nailed it by saying that increased weight will require a higher angle of attack.
A heavier plane will indeed require a higher angle of attack to provide more lift.
You mentioned that the plane is flying horizontally, with a constant speed. In that case, the plane is at an equilibrium, the weight is compensated by the lift and drag is compensated by traction.
Let's say we want to fly the same plane, but with more cargo. We would need to increase lift. What can we act upon ? The lift formula sums it all :

Rz lift , ρ Volumic mass of air, S wing surface, V Speed, Cz Lift coefficient
Therefore a weight increase will either require to increase the volumic mass of air (huh?), the wing surface, the speed of the plane or the lift coefficient.

to increase the density of air, just fly lower. For extreme density, try Siberia in winter.


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How is flight envelope protection implemented in Dassault Falcons?
In normal laws, envelope protection is implemented differently on Boeing and Airbus fly-by-wire systems, as highlighted in this question. I don't want to reopen a debate about which one is better (both have been proven to be good), I'm looking for information about another manufacturer's implementation.
If I understand correctly, Airbus won't allow the pilot to exceed limitations (such as load factor) while Boeing will make them really hard to exceed them.
Dassault Falcon jets (after the 7X) are fitted with fly-by-wire implementing flight envelope protections (as mentioned in this article.
I would like to know what happen if the pilot's input tend to exceed flight envelope limits (high-G maneuver -both negative and positive G-, high angle of attack, over-speed, high bank angle).
On a fully functional Dassault Falcon 7X (or newer) in normal laws, can the pilot put the aircraft outside its flight envelope? Can the pilot make the aircraft stall (excessive AoA), perform high-G maneuver (excessive load factor), put the aircraft in inverted flight (bank angle > 90°)? Will the aircraft refuse pilot's input or just make them hard ? In short, how is flight envelope protection implemented on Dassault falcon jets?

My information on the Falcon 7x and 8x comes from Aviation Week and Flying Magazine pilot-reports.
It would appear that the Dassault FBW philosophy is very closely aligned with that of Airbus. Namely,

It employs a C* control law without speed stability;
It uses passive sidestick instead of control column;
It employs 3-axis envelope protection, complete with bank angle protection (via positive spiral stability beyond 35 deg), load factor protection, AOA protection and high speed protection;
It even has a control-by-wire backup system similar to the mechanical backup available in the A320 and A330.

The fact that it has a C* control law, which provides flight path stability and not speed stability, and that it has a passive sidestick without force feedback, the envelope protection would appear to be a must-have for certification, rather than nice-to-have. By all indications, the limitations are hard constraints that cannot be overridden by pilot inputs in Normal Law.

It's really strange that Dassault FBW documentation is so hard to find.

@ManuH I don't think any biz jet documentation is easy to come by :) For FBW, you may also want to check out the Gulfstream vs Bombardier vs Embraer philosophies, each of them has different takes

The 7X in normal law will attenuate sidestick inputs to prevent excessive AOA (getting too close to the stall), or excessive load factors.
As a practical example (I've done this in the 7X simulator, not in the actual aircraft), if in level flight you pull the sidestick quickly back past the soft stop, all the way to the hard stop and hold it there, the nose will initially rise, but the G-loading will not exceed limits. As the 7X then approaches stalling angle, the aircraft will start to pitch down - at a fairly gentle rate - even if you hold the stick full aft, to prevent getting too close to the stall.
During our sim training, the instructor did something very interesting: Made us fly a 63-degree bank turn (we HAD to keep it above 60 degrees) with slats and flaps extended, at 300 feet AGL. This requires constant fine bank inputs to the sidestick, as the aircraft is always trying to roll back to 30 degrees. The flap configuration meant that the G limit was now +2.0.
Well, it was impossible to maintain level flight! Of course, perfectly logical, since you would need to pull more than 2G to stay level. But I'm still not sure how I feel about this "protection feature".
I agree that there is practically zero documentation about the Dassault or any other FBW systems. Might be due to FBW's military origins...
But, it would be nice to have a bit more info. A hard-landing accident on an Airbus was investigated, and if memory serves me well it revealed that below a height of 50 feet their FBW system changes in the way it reacts. I was very curious to find out if Dassault also does this, but never found anything in writing.


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How do I factor sweep and dihedral in the lift equation?
I'm guessing that you add cos(angle of dihedral) to calculate lift loss by dihedral, but I'm not sure how to factor in the lift loss by the wing swept. Could someone help me include lift loss due to dihedral and sweep? Thank you
Note: I mean without factoring it into the lift coefficient.

I don't think you can avoid 'factoring it [wing sweep] into the lift coefficient'. The sweep changes the angle of attack response (reducing the lift slope in the first place, but not only that), rather than lift per se. If you linearise though, then it's possible.

Snorri Gudmundsson's "General Aviation Aircraft Design"'s equation 9-57 allows you to estimate the lift coefficient curve's linear slope for trapezoidal/swept wings:

Note: the equation is in radians, not degrees
As for the dihedral, Equation 9-33 estimates the impact on angle of attack and Equation 9-35 estimates the resulting lift force:

Let me know if it's not what you wanted, though, I didn't quite get the "without factoring it into the lift coefficient." part - isn't the question's title about how it impacts the lift coefficient?


text:
How can a designer deal with a strict requirement when designing an aircraft?
I'm working on designing an aircraft with some specific requirements (Vstall, Vmax, Ceiling, Payload, Range, RoC, Slanding/takeoff etc).
I'm following Anderson's and Raymer's ways (trying to find the best for me).
Right now I'm stuck at the step of estimating for the first time the Wing Loading (W/S). Doing the calculations about the W/S constraints (Vstall, Landing distance from Anderson) I found that the Stall constraint is very strict. The result that I get is a W/S = 24 lb/ft2 for a twin turboprop with a estimated gross takeoff weight of 16k lbs. The constraint of landing distance gives a W/S of 62.
Looking at historical data (and some similar aircraft) and Raymer's/Anderson's tables the W/S = 24 is way off the ordinary values. I do understand that it is just an estimation.
What would you suggest to do so I can satisfy the requirement and be able to get a higher W/S?
I thought about changing the flaps that I used (Split Flap) but I don't know if it's possible and realistic to have "Double Slotted Flaps and Slats" for a simple twin turboprop that has a simple mission.
edit: After the comment, here are the equations about the Vstall and Landing Distance.

and

No VtC from me, but this may be far too broad. I'm not an expert, though, and could be totally wrong...

Can you list the equations that derive the "stall constrained W/S" and "T/O constrained W/S"?

@JZYL Check the post, I edited it.

This smells a lot like the wrong c${L{max}}$ is used for the stall speed. Are you using the clean c${L{max}}$ when the task asks for the c${L{max}}$ in landing configuration? The formulas are fine and dandy, but what numbers do you use for the parameters? Single slotted flaps with a bit of rearwards extension would be adequate, but the choice really depends on the requirements. Maybe that specific combination can only be fulfilled with double slotted Fowler flaps and slats. A split flap is a bit old-fashioned, don't you think?

@PeterKämpf My guess is that the Vstall is very small. Actually the required Vstall is 100kmh. I follow Anderson's way for the Cl: I selected airfoils for my wing and got their Clmax, added the extra Cl that the specific flaps add (+1.2 according to Anderson) and calculated the 3D effects of the induced drag with the simplified CL = 0.9*Cl for a wing with AR>5. My CLmax equals 2.6. I guess yes I could use better flaps.
The aircraft has a required range of 1000km, Vmax 500kph, ceiling 10km. I think that 100kph as Vstall is low. I would say 150kph is a better "standard" as fas as I checked.

I agree, 100 km/h is ridiculously low when v$_{max}$ is supposed to be 500 km/h. Even light GA aircraft may have 113 km/h stall speed and for a twin with 16,000 lbs this limit is much higher. Take the Metro III (similar gross weight): It only flies 459 km/h but has a stall speed of 87 kts (= 161 km/h) with flaps down. Both go together: A low stall speed means a low cruise speed and vice versa. Those requirements don't match.

So here's the question, usually Vstall is needed to be low for takeoff and landing performance. What's the point of a very low stall speed if it will not improve field performance?

@PeterKämpf Since I dont have the experience of designing a real aircraft, would it be possible to negotiate with the "customer" (who decided the requirements) to change the Vstall to a higher value? Or should I find a way around it? (Somehow increasing the CLmax). Continuing the design process with such a low W/S what would be the impact later? Sorry for all those questions.

@ChrisP: Keeping the low stall speed means you need to carry around a huge wing which creates more drag at cruise speed, so the engines need to get bigger. And the fuel tanks. Which in turn requires a bigger wing. Repeat. The result is an economic disaster for the customer. Of course, with complex flaps and slats you can raise the lift coefficient, but still this will result in a big and heavy wing which creates more drag than necessary. In short, renegotiate that particular requirement because it will lead to absurd results. Could it be your supervisor has even less experience than you?


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What radial is the aircraft crossing?
I'm working through the ASA Private Pilot 2020 Test Prep questions and there's a question I don't understand the answer for.

3579 (Refer to Figure 28, illustration 5) The VOR receiver has the
indication shown. What radial is the aircraft crossing?

A - 030
B - 210
C - 300

I thought the answer was B - 210 but that is the heading, not the crossing.
The correct answer is A - 030, but I'm not sure why. What does it mean to be crossing a radial?

210 is NOT the heading. That instrument has no heading information available to the pilot. It only allows the pilot to select a desired track to or from a VOR station.

A radial is always pointing (radiating) away from the VOR. But your instrument shows the TO/FROM indicator as TO. So you are crossing a 210 course towards the VOR, which means you are on the 030 radial from the VOR.

What does it mean to be crossing a radial?

This just refers to the fact that you cannot determine your heading (or track) from the CDI. Based on this instrument alone, you could be following the track to VOR right now or just happen to cross the radial at this moment (but actually fly in a totally different direction).

Another term (not necessarily an aviation term) for this would be the reciprocal bearing.

Radials radiate outwards from a station, just like the magnetic bearings on a compass rose.  So, the outward radial is just the reciprocal of the inbound bearing.
The key is the “TO” flag.  If you were to be flying a heading of 210 degrees you would fly TO the navaid.
Draw it out on a piece of paper, or refer to a chart or your study material.  Orientation questions like this benefit most from visuals.
Also, don't get confused thinking this instrument is a gyroscopic compass card.  It is not slaved to any inputs, the course showing on the face is manually dialed in and remains as set until you change it.
“Crossing a radial” simply means that you are not tracking along it either inbound or outbound to/from the navaid.  (You are going across it at some angle.)  This is a useful technique for identifying intersections if you have two VOR receivers:  With one dialed in to the radial you are tracking, set the other one to the other VOR station that defines the fix, and dial in that radial on your OBS.  When the CDI bar centers up it means you are crossing that radial.